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A063747
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Sign of n-th coefficient of power series for 1/Gamma(1-x) where Gamma is the Gamma function.
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1
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1, -1, -1, 1, 1, 1, -1, -1, -1, 1, 1, 1, -1, -1, -1, -1, 1, 1, 1, -1, -1, -1, -1, 1, 1, 1, 1, -1, -1, -1, 1, 1, 1, 1, -1, -1, -1, -1, 1, 1, 1, 1, 1, -1, -1, -1, -1, 1, 1, 1, 1, -1, -1, -1, -1, 1, 1, 1, 1, 1, -1, -1, -1, -1, 1, 1, 1, 1, -1, -1, -1, -1, -1, 1, 1
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OFFSET
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0,1
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COMMENTS
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Computing this sequence was discussed on the seqfan mailing list in May 2023. Typical floating-point precision is not sufficient to compute more than a few terms of this sequence. Therefore, either a high precision needs to be set; e.g, a precision of round(n*log(n)/Pi) [see formula (3.21) in Fekih-Ahmed], or the computation must be done with an arbitrary precision method (e.g. interval arithmetic, FLINT, or the Java below). - Sean A. Irvine, May 21 2023
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LINKS
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EXAMPLE
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1/Gamma(1-x) = 1 - gamma*x - 0.6558...*x^2 + 0.042...*x^3 + 0.1665...*x^4 + 0.0421...*x^5 - 0.0096..*x^6 +.... hence sequence begins 1,-1,-1,1,1,1,-1,...
[x^46] 1/Gamma(1-x) = -4.445829736550756882101590352124643637401436685748718288670...*10^-39, from which a(46)=-1. Sean A. Irvine, May 21 2023
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MATHEMATICA
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precis = 200;
g[1, k_] := g[1, k] = EulerGamma^k/k!;
g[m_, k_] := g[m, k] = Zeta[m]^k/(k! m^k) // N[#, precis]&;
f[0, 0] = 1; f[n_ /; n > 0, 0] = 0;
f[n_, m_] := f[n, m] = Sum[g[m, k]*f[n - k m, m - 1],
{k, 0, n/m}] // N[#, precis]&;
c[i_] := f[i, i]; b[0] = 1;
b[i_] := b[i] = -Sum[c[j]*b[i - j], {j, 1, i}];
a[n_] := Sign[b[n]];
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PROG
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(PARI) a(n)=sign(polcoeff(1/(gamma(1-x +O(x^(n+1)))), n))
(PARI) my(x='x+O('x^100)); apply(sign, Vec(1/(gamma(1-x)))) \\ Michel Marcus, May 19 2023
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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